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2 years ago

14

Peardeck - slide #21 - 7th-grade math - fill in the chart of negative and positive integers

1 answer:

Lunna [17]2 years ago

8 0

Answer:

lllllllllooooooooooollllllooolloollllll

Step-by-step explanation:

alkdndnbdnd

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!!20 points and Brainliest please help!!!

vitfil [10]

D) ac + ad + bc + bd

Hope it helps :)

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2 years ago

Read 2 more answers

Find the zeros of the function by first factoring the polynomial.

torisob [31]

Step-by-step explanation:

y2 + 4y - 5

( y + 4 )( y - 1 )

so you just need to break down the sum

5 0

2 years ago

If you have a cubic polynomial of the form y = ax^3 + bx^2 + cx + d and lets say it passes through the points (2,28), (-1, -5),

nikklg [1K]

Step-by-step explanation:

<u>Step 1: Solve using the first point</u>

<em>(2, 28)</em>

$28 = a(2)^3 + b(2)^2 + c(2) + d$

$28 = 8a + 4b + 2c + d$

<u>Step 2: Solve using the second point</u>

<em>(-1, -5)</em>

$-5 = a(-1)^3 + b(-1)^2 + c(-1) + d$

$-5 = -a + b - c + d$

<u>Step 3: Solve using the third point</u>

<em>(4, 220)</em>

$220 = a(4)^3 + b(4)^2 + c(4) + d$

$220 = 64a + 16b + 4c + d$

<u>Step 4: Solve using the fourth point</u>

<em>(-2, -20)</em>

$-20 = a(-2)^3 + b(-2)^2 + c(-2) + d$

$-20 = -8a + 4b - 2c + d$

<u>Step 5: Combine the first and fourth equations</u>

<u /> $28 - 20 = 8a - 8a + 4b + 4b + 2c - 2c + d + d$

$8 = 8b + 2d$

$8 - 8b = 8b - 8b + 2d$

$(8 -8b)/2 = 2d/2$

$4 - 4b = d$

<u>Step 6: Solve for c in the second equation</u>

$-5 + 5 = -a + b - c + d + 5$

$0 + c = -a + b - c + c + d + 5$

$c = -a + b + d + 5$

<u>Step 7: Substitute d with the stuff we got in step 5</u>

$c = -a + b + (4 - 4b) + 5$

$c = -a + b + 4 - 4b + 5$

$c = -a - 3b + 9$

<u>Step 8: Substitute d and c into the first equation</u>

<u /> $28 = 8a + 4b + 2(-a - 3b + 9) + (4 - 4b)$

$28 = 8a + 4b - 2a - 6b + 18 + 4 - 4b$

$28 - 22 = 6a - 6b + 22 - 22$

$6 / 6 = (6a - 6b) / 6$

$1 + b = a - b + b$

$1 + b = a$

<u>Step 9: Substitute a, b, and c into the third equation</u>

$220 = 64(1 + b) + 16b + 4(-(1 + b) - 3b + 9) + (4 - 4b)$

$220 = 64 + 64b + 16b + 4(-1 - b - 3b + 9) + 4 - 4b$

$220 - 100 = 60b + 100 - 100$

$120 / 60 = 60b / 60$

$2 = b$

<u>Step 10: Find a using b = 2</u>

$a = b + 1$

$a = (2) + 1$

$a = 3$

<u>Step 11: Find c using a = 3 and b = 2</u>

$c = -a - 3b + 9$

$c = -(3) - 3(2) + 9$

$c = -3 - 6 + 9$

$c = 0$

<u>Step 12: Find d using b = 2</u>

$d = 4 - 4b$

$d = 4 - 4(2)$

$d = 4 - 8$

$d = -4$

Answer: $a = 3, b = 2, c = 0,d = -4$

6 0

2 years ago

Read 2 more answers

Solve, using linear combination. 5x + y = 2 4x + y = 4

mylen [45]

I hope this helps you

3 0

2 years ago

Consider the equation 5x^2-10x+c=0. What values of c result in the equation having a complex solutions? Represent your answer on

stiv31 [10]

Answer:

When we have a quadratic equation:

a*x^2 + b*x + c = 0

There is something called the determinant, and this is:

D = b^2 - 4*a*c

If D < 0, then the we will have complex solutions.

In our case, we have

5*x^2 - 10*x + c = 0

Then the determinant is:

D = (-10)^2 - 4*5*c = 100 - 4*5*c

And we want this to be smaller than zero, then let's find the value of c such that the determinant is exactly zero:

D = 0 = 100 - 4*5*c

4*5*c = 100

20*c = 100

c = 100/20 = 5

As c is multiplicating the negative term in the equation, if c increases, then we will have that D < 0.

This means that c must be larger than 5 if we want to have complex solutions,

c > 5.

I can not represent this in your number line, but this would be represented with a white dot in the five, that extends infinitely to the right, something like the image below:

8 0

2 years ago